Codeforces 901C Bipartite Segments

题目蓝链

Solution

因为这是一个仙人掌图，所以我们可以通过(tarjan)缩点双找出图中所有的简单环。然后我们找出每一个环中的最大编号和最小编号，并按最大编号从小到大排序。然后我们把询问离线下来，并且挂在询问的右端点处

然后我们从左往右扫，处理每一个节点处的询问之前，先把最左能走的不会形成环的位置更新出来。然后用线段树把这一段区间加(1)

询问就直接再线段树上区间求和就行了

还有一种做法，就是直接预处理出每一个点最左能走的不会形成环的位置，记为(pos[i])。对于查询([l, r])，二分找到最小的(i)使得(pos[i] > l)。然后我们分别对([l, i - 1])和([i, r])计算贡献，加起来就可以了

Code

#include <bits/stdc++.h>

using namespace std;

#define fst first
#define snd second
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

inline int read() {
	int sum = 0, fg = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
	for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
	return fg * sum;
}

const int maxn = 3e5 + 10;

vector<int> g[maxn];
vector<pii> v[maxn];

int d[maxn], low[maxn], pre[maxn], dfn;
int S[maxn], top, cnt, n, m, q;

pii p[maxn];
LL ans[maxn];

void tarjan(int now, int f) {
	d[now] = low[now] = ++dfn;
	S[++top] = now;
	for (int i = 0; i < g[now].size(); i++) {
		int son = g[now][i];
		if (son == f) continue;
		if (!d[son]) {
			tarjan(son, now);
			low[now] = min(low[now], low[son]);
			if (low[son] >= d[now]) {
				int Max = 0, Min = n + 1, tot = 0;
				while (top) {
					int x = S[top--];
					Max = max(Max, x);
					Min = min(Min, x);
					++tot;
					if (x == son) break;
				}
				Max = max(Max, now), Min = min(Min, now);
				if (tot > 1) p[++cnt] = (pii){Max, Min};
			}
		}else low[now] = min(low[now], d[son]);
	}
}

namespace ST {

	LL A[maxn << 2], tag[maxn << 2];
#define ls (rt << 1)
#define rs (rt << 1 | 1)

	void push_up(int rt) { A[rt] = A[ls] + A[rs]; }

	void push_down(int rt, int l, int r) {
		if (!tag[rt]) return;
		int mid = (l + r) >> 1;
		tag[ls] += tag[rt];
		tag[rs] += tag[rt];
		A[ls] += tag[rt] * (mid - l + 1);
		A[rs] += tag[rt] * (r - mid);
		tag[rt] = 0;
	}

	void change(int rt, int l, int r, int L, int R) {
		if (L <= l && r <= R) {
			A[rt] += r - l + 1, ++tag[rt]; return;
		}
		push_down(rt, l, r);
		int mid = (l + r) >> 1;
		if (L <= mid) change(ls, l, mid, L, R);
		if (R > mid) change(rs, mid + 1, r, L, R);
		push_up(rt);
	}

	LL query(int rt, int l, int r, int L, int R) {
		if (L <= l && r <= R) return A[rt];
		push_down(rt, l, r);
		int mid = (l + r) >> 1; LL res = 0;
		if (L <= mid) res += query(ls, l, mid, L, R);
		if (R > mid) res += query(rs, mid + 1, r, L, R);
		return res;
	}

}

int main() {
#ifdef xunzhen
	freopen("graph.in", "r", stdin);
	freopen("graph.out", "w", stdout);
#endif

	n = read(), m = read();
	for (int i = 1; i <= m; i++) {
		int x = read(), y = read();
		g[x].push_back(y);
		g[y].push_back(x);
	}

	q = read();
	for (int i = 1; i <= q; i++) {
		int x = read(), y = read();
		v[y].push_back((pii){x, i});
	}

	for (int i = 1; i <= n; i++) if (!d[i]) tarjan(i, 0);

	sort(p + 1, p + cnt + 1);
	int j = 1, Max = 0;
	for (int i = 1; i <= n; i++) {
		while (j <= cnt && p[j].fst <= i) Max = max(Max, p[j++].snd);
		ST::change(1, 1, n, Max + 1, i);
		for (int j = 0; j < v[i].size(); j++)
			ans[v[i][j].snd] = ST::query(1, 1, n, v[i][j].fst, i);
	}

	for (int i = 1; i <= q; i++) printf("%lld
", ans[i]);

	return 0;
}

Summary

考试的时候没有注意到这是一个仙人掌，以为这是一个仙人球，然后我就打了一个缩边双...